# Proof:$X_n\overset{a.s}{\rightarrow}X,Y_n\overset{a.s}{\rightarrow}Y \Rightarrow X_n+Y_n\overset{a.s.}{\rightarrow}X+Y$

I have a simple question but i didn't find a full solution to it anywhere.
I would like to know if my demonstration of the following theorem is correct.
Property:
Let $$X_n\left ( \omega \right )\overset{a.s.}{\rightarrow}X\left ( \omega \right )$$ and $$Y_n\left ( \omega \right )\overset{a.s.}{\rightarrow}Y\left ( \omega \right )$$ on $$(\Omega ;F;\mathbb{P})$$ a probability space. So $$X_n\left ( \omega \right ) + Y_n\left ( \omega \right )\overset{a.s.}{\rightarrow}X\left ( \omega \right )+Y\left ( \omega \right )$$

Proove:
$$S_1=\left \{ \omega \in \Omega:X_n(\omega)\overset{p.w.}{\rightarrow} X(\omega) \cap Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \}\subseteq \left \{ \omega \in \Omega:Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \}$$. Furthermore $$\mathbb{P}( \left \{ \omega \in \Omega \; : \; X_n(\omega) \overset{p.w.}{\rightarrow}X(\omega) \right \})=1 > \mathbb{P}( \left \{ \omega \in \Omega:Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \})=0$$ by def. So: $$\mathbb{P}(S_1)=0$$.
$$S_2=\left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \cap Y_n(\omega)\overset{p.w.}{\rightarrow} Y(\omega) \right \}\subseteq \left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \right \}$$. So for the same reason as before $$\mathbb{P}(S_2)=0$$
$$S_3=\left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \cap Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \}\subseteq \left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \right \}$$ So for the same reason as before $$\mathbb{P}(S_3)=0$$

Now by definition the set of all $$\omega \in \Omega$$ where $$X_n(\omega)+Y_n(\omega)\overset{not \; p.w. \; cvge}{\rightarrow}X(\omega)+Y(\omega)$$, is include in $$S_1\cup S_2\cup S_3$$ (as every where $$X_n$$ and $$Y_n$$ all ready p.w. converge we have $$X_n+Y_n$$ p.w. too).
Hence: $$\mathbb{P}(\omega \in \Omega : X_n(\omega)+Y_n(\omega)\overset{not \; p.w. \; cvge}{\rightarrow}X(\omega)+Y(\omega))\leq \mathbb{P}(S_1 \cup S_2 \cup S_3)\leq \mathbb{P}(S_1)+\mathbb{P}(S_2)+\mathbb{P}(S_3)=0$$
So we just proved that the measure of the subset of $$\Omega$$ where there is not p.w. convergence of $$X_n+Y_n$$ has a measure of zero and, hence, the measure of the set where there is p.w. convergence has a size of 1 (by measure property). And this is exactly the defintion of $$X_n(\omega)+Y_n(\omega)\overset{a.s.}{\rightarrow} X(\omega)+Y(\omega)$$. Q.E.D

Is this correct?
• This is correct as far as I can tell, but you seem to have gone about it in a somewhat round about way. You know that, by definition, the set of $\omega$ such that $Y_n(\omega)$ does not converge pointwise has measure zero. Similarly for $X_n$. The intersection of zero measure sets has measure zero, assuming the intersection is measurable. Since the points where the sum does not converge pointwise is a subset of the points where at least one fails to converge pointwise (and presumably the measure is complete), the set on which convergence fails has measure zero Commented Sep 1, 2022 at 21:35
• If i understood you well you use the fact that $A\cap B \subseteq A$ and that too we have $A\cap B \subseteq B$ . This is exactly what i am using, i fully explained this idea only for $S_1$ Commented Sep 1, 2022 at 21:38
The idea is sound, although there is a minor typo towards the end $$\mathbb{P}(X_n+Y_n \text{ does not converge})\le\mathbb{P}(S_1 \cup S_2 \cup S_3)$$ instead of intersections like you have used.
In any case, you've chosen to prove that the set of divergence has measure zero which is fine. There is a slightly different approach that may be a little shorter. If we let $$S_1=\{X_n \text{ converges}\}$$ and $$S_2=\{Y_n \text{ converges}\}$$ then $$\mathbb{P}(S_1)=\mathbb{P}(S_2)=1$$. Thus, $$\mathbb{P}(S_1\cap S_2)=1$$ and on $$S_1\cap S_2$$, we know from elementary real analysis that $$X_n+Y_n$$ converges.